Computing the odds that max. $11$ items are late, if $132$ items are delivered with $96.2\%$ success rate per item. Why two methods give two results?

Question

Probability of success (p) = 96.2% (Item delivered on time)

Number of repetitions (n) = 132 (Items that has to be delivered)

Question: What is the probability of max. 11 items being late.

Note: Both of these methods are valid ways of calculating the probability, so why are the outcomes not equal?

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METHOD 1

X: Number of items that arrive on time.

X~b(132, 96.2%)

I calculate the probability of Max 121 (=132-11) arrive on time and subtract it from 1 (using Npsire), which must give the probability of max 11 items being late.

$$1-P(X \leq121)=1-\text{binomCdf}(132,96.2\text{%},121) \rightarrow 0.987732759841$$

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METHOD 2

X: Number of items that are late.

X~b(132, 3.8%) (3.8%=100%-96.2%)

I calculate the probability of max. 11 are late.

$$P(x \leq 11)=\text{binomCdf}(132,3.8\text{%},11) \rightarrow 0.995336140422$$

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COMPARISON

$$1-P(X \leq121)=1-\text{binomCdf}(132,96.2\text{%},121) \rightarrow 0.987732759841$$ $$P(x \leq 11)=\text{binomCdf}(132,3.8\text{%},11) \rightarrow 0.995336140422$$

$$0.987732759841 \not= 0.995336140422$$

What is going on?

Answer 1

The difference of $0.9953-0.9877=0.0076$ is the probability that exactly $121$ items are delivered on time and $11$ late, which is ${132 \choose 121} 0.962^{121} 0.038^{11} \approx 0.0076$

Your first calculation includes the probability of this event in its sum and so wrongly excludes this from the final probability, thus failing to answer the question as asked

Your second calculation includes the probability of this event in its sum and so correctly includes this in the final probability, thus succeeding in answering the question

Answer 2

Let $X$ and $Y$ be the numbers of items arriving on time and late, respectively.

Then: $X+Y=132 \Rightarrow Y=132-X$.

So: $$\mathbb P(Y\le 11)=1-P(Y\ge 12)=1-P(132-X\ge 12)=1-P(X\le 120)=0.99533614042\color{red}9...$$ Reference: WA answer.